Suppose the compact Riemann surface $C$ is an n-fold branched cover of $\mathbb{P}^1$ branched at exactly four points $x_1,x_2,x_3$ and $x_4$. I believe that $C$ is a smooth genus $(n-1)$ Riemann surface. Is this true?

In case 1 is true, what is the algebraic equation for this Riemann surface? [Naively, I thought $y^n=(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ would do the job, but this surface is singular when it is given as a sub-variety of $\mathbb{P}^2$. Do I have to embed the initial $\mathbb{P}^1$ in a more complicated weighted projective space? How do I resolve the singularity to get a smooth genus $(n-1)$ surface?]

Given the algebraic equation for the Riemann surface $C$ of question 2, what are the $(n-1)$ holomorphic differentials?

I have a basic knowledge of Riemann surfaces obtained through undergraduate/early graduate level courses, but would like to learn more. In particular I would like to be address problems similar to questions 1, 2, 3 on my own with relative ease. Do you have any references(on/offline) that I can learn from?

2 Answers
2

The formula is true if the cover is branching is "total", that is, if all sheets of the cover join into a single point. This is a straightforward application of the Riemann-Hurwitz formula (algebraic geometry). However, as Jason Starr pointed out, if only some of the sheets join, the genus will be lower - another straightforward application of this formula.

One way to get the correct surface is to blow up the singular point, which involves looking at $\mathbb P^2 \times \mathbb P^1$. Are you familiar with this process? Since your formula gives the correct birational equivalence class, blowing-up will give the correct surface.

If you get one holomorphic differential, the other ones will be rational function multiples of it. You could figure out where it has zeroes and try to find enough rational functions that have poles there. But this probably isn't the best way. Another, somewhat silly, way is that if you know an embedding into a high-enough dimensional space, you can get all the holomorphic differentials as restrictions of holomorphic differentials on $\mathbb P^n$. You can get bigger-dimensional embeddings out of smaller ones by using the Veronese maps.

This is fairly standard algebraic geometry, so I would suggest a standard algebraic geometry text like Hartshorne. However learning algebraic geometry is quite a bit of work just to answer these questions. I don't know good sources on it from a complex analytic perspective.

Even in the "totally branched" case there's more than one cover once $n>2$, none of which is $y^n = (x-x_1) (x-x_2) (x-x_3) (x-x_4)$ unless $n=4$ (lest the point at infinity also ramify). Three examples are $y^n = ((x-x_1) (x-x_2)) / ((x-x_3) (x-x_4))$ and its permutations.
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Noam D. ElkiesJan 4 '12 at 4:18

For 4., take a look at Rick Miranda's `Algebraic curves and Riemann surfaces'.
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Chris BravJan 4 '12 at 12:43

Thank you for the responses! Just an additional question on the remark that there are multiple covers once $n>2$. In that case, would all the different choices of covers have equivalent complex structures/period matrices? Thanks again!
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D. S. ParkJan 5 '12 at 1:16

1

@D. S. Park: The answer to your new question is also "no". Essentially you are asking if one of these cyclic $\mathbb{Z}_n$-covers $X$ could have a second (inequivalent) structure of $\mathbb{Z}_n$-cover. But then the subgroup $G$ of $\text{Aut}(X)$ generated by these two copies of $\mathbb{Z}_n$ would be strictly larger than $\mathbb{Z}_n$ (coming from one of the two structures). The quotient of $X$ by $G$ would be a curve which is dominated by $X/\mathbb{Z}_n$, which is already $\mathbb{P}^1$. Thus $X/G$ is $\mathbb{P}^1$. continued
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Jason StarrJan 5 '12 at 17:42

@D. S. Park: Continued. For every branch point $b$ of $X\to X/G$, the ramification type of every preimage $x$ in $X$ is the same. Consider the map $X/\mathbb{Z}_n\to X/G$. The degree of this map, $d$, equals the index of $\mathbb{Z}_n$ in $G$. By Riemann-Hurwitz, the degree of the branch divisor is $2d-2$. But this quickly gives a contradiction unless the map is totally branched over 2 points. continued.
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Jason StarrJan 5 '12 at 17:57

Look at the Riemann-Hurwitz [RH] formula carefully.
I recommend Plane Algebraic Curves by G. Fisher, or Complex Algebraic Curves
by Kirwan. All kinds of interesting things can happen, and in symmetric ways,
if 4 divides n. For example, you might try to have the inverse image of each
of branch point be exactly 3 points, and look for the local branching at each branch point to be 2: as in $z \to z^2$. The total degree will then be $6 = 2*3$.
Does the Riemann surface exist? If so, its genus must be 3, by RH.